Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 44644

The numbers D, N, and P are positive integers, such that D < N, and N [#permalink]
Show Tags
17 Mar 2015, 05:58
2
This post received KUDOS
Expert's post
20
This post was BOOKMARKED
Question Stats:
23% (02:13) correct 77% (02:02) wrong based on 259 sessions
HideShow timer Statistics



Manager
Joined: 14 Sep 2014
Posts: 106
Concentration: Technology, Finance
WE: Analyst (Other)

Re: The numbers D, N, and P are positive integers, such that D < N, and N [#permalink]
Show Tags
17 Mar 2015, 07:51
3
This post received KUDOS
I will say C. Please let me know if I am missing something.
(1) The four factors of N include 1, D, something, and N. Therefore, N is the product of D and something. Test case 1: N = 6 so D is either 2 or 3 and therefore is prime. Test case 2: N = 8 so D is 4 and not prime (D cannot be 2 because 8 is a power of 2, which violates the constraints.)
(2) Possible values for D include 5, 11, 29, 83, 245, ... etc. We have prime and nonprime possibilities, so this is insufficient.
Combining the statements, D must be one of the prime values given in statement 2, otherwise N would have more than four factors.



Director
Joined: 07 Aug 2011
Posts: 567
Concentration: International Business, Technology

Re: The numbers D, N, and P are positive integers, such that D < N, and N [#permalink]
Show Tags
17 Mar 2015, 08:10
Bunuel wrote: The numbers D, N, and P are positive integers, such that D < N, and N is not a power of D. Is D a prime number?
(1) N has exactly four factors, and D is a factor of N (2) D = (3^P) + 2
Kudos for a correct solution. Option A:N has exactly four factors, and D is a factor of N : Not Sufficient \(2^1 * 3^1\) factors are 1,2,3,6 ; D can be 2 or 1 . Not sufficient. \(19^1 * 2^1\) factors are 1,2,19,38 ; D can be 19,2,1 . (note 1 is not prime) Option B: D = (3^P) + 2 Not Sufficient \(P=1; D=5 is prime ? Yes\) \(P=5; D=245 is prime ? NO\) A && B from B we know that D is and ODD Integer and from A we have D can be 19,2,1 . (note 1 is not prime) 19 is prime but 1 is not so still the statement is insufficient. Answer E
_________________
Thanks, Lucky
_______________________________________________________ Kindly press the to appreciate my post !!



Manager
Joined: 14 Sep 2014
Posts: 106
Concentration: Technology, Finance
WE: Analyst (Other)

The numbers D, N, and P are positive integers, such that D < N, and N [#permalink]
Show Tags
17 Mar 2015, 10:02
1
This post received KUDOS
Lucky2783 wrote: D can be 19,2,1 . (note 1 is not prime) 19 is prime but 1 is not so still the statement is insufficient. If D = 1, then 3^P = 1 and P is not a positive integer, violating the constraint given in the problem. (Same with D = 2 or 19)



Director
Joined: 07 Aug 2011
Posts: 567
Concentration: International Business, Technology

Re: The numbers D, N, and P are positive integers, such that D < N, and N [#permalink]
Show Tags
17 Mar 2015, 10:24
sterling19 wrote: Lucky2783 wrote: D can be 19,2,1 . (note 1 is not prime) 19 is prime but 1 is not so still the statement is insufficient. If D = 1, then 3^P = 1 and P is not a positive integer, violating the constraint given in the problem. (Same with D = 2 or 19) good catch !! Answer is C.
_________________
Thanks, Lucky
_______________________________________________________ Kindly press the to appreciate my post !!



Math Expert
Joined: 02 Sep 2009
Posts: 44644

The numbers D, N, and P are positive integers, such that D < N, and N [#permalink]
Show Tags
23 Mar 2015, 05:57
2
This post received KUDOS
Expert's post
6
This post was BOOKMARKED
Bunuel wrote: The numbers D, N, and P are positive integers, such that D < N, and N is not a power of D. Is D a prime number?
(1) N has exactly four factors, and D is a factor of N (2) D = (3^P) + 2
Kudos for a correct solution. MAGOOSH OFFICIAL SOLUTION:This is a tricky one. Statement #1: two kinds of numbers have exactly four factors: (a) products of two distinct prime numbers, and (b) cubes of prime numbers. The product of two distinct prime numbers S and T would have factors {1, S, T, ST}. For example, the factors of 10 are (1, 2, 5, 10), and the factors of 21 are {1, 3, 7, 21}. The cube of a prime number S would have as factors 1, S, S squared, and S cubed. For example, 8 has factors {1, 2, 4, 8} and 27 has factors {1, 3, 9, 27}. We know N is not a power of D, so the second case is excluded. N must be the product of two distinct prime numbers. We know D < N, so of the four factors, D can’t be the product of the two prime numbers. D could be either of the prime number factors, or D could be 1, which is not a prime number. Because D could either be a prime number or 1, we cannot give a definitive answer to the question. This statement, alone and by itself, is not sufficient. Statement #2: this is tricky. The first few plugins seem to reveal a pattern. Even if you sense a pattern, it’s important to remember that plugging in numbers alone is never enough to establish that a DS statement is sufficient. Here, if we persevered to one more plugin, we would find the one that breaks the pattern. P = 5 > 3^5 + 2 = 243, which is not a prime. That gives another answer to the prompt, so we know this statement is not sufficient. To avoid a lot of plugging in, it’s also very good to know that in mathematics, prime numbers are notorious for not following any easy pattern. It is impossible to produce an algebraic formula that will always produce prime numbers. In fact, this is more than you need to know, but the hardest unsolved question in higher mathematics, the Riemann Hypothesis, concerns the pattern of prime numbers; mathematicians have been working on this since 1859, and no one has proven it yet. Suffice to say that no oneline algebraic formula is going to unlock the mystery of prime numbers! Combined statements: according to the information in statement #1, either D = 1 or D is a prime number. Well, statement #2 excludes the possibility that D = 1, because that number cannot be written as two more than a power of 3. Therefore, D must be a prime number. We have a definitive answer to the prompt question. Combined, the statements are sufficient. Answer = (C) Attachment:
ghdmpp_img25.png [ 6.41 KiB  Viewed 3244 times ]
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Director
Joined: 07 Aug 2011
Posts: 567
Concentration: International Business, Technology

Re: The numbers D, N, and P are positive integers, such that D < N, and N [#permalink]
Show Tags
23 Mar 2015, 06:50
Lucky2783 wrote: sterling19 wrote: Lucky2783 wrote: D can be 19,2,1 . (note 1 is not prime) 19 is prime but 1 is not so still the statement is insufficient. If D = 1, then 3^P = 1 and P is not a positive integer, violating the constraint given in the problem. (Same with D = 2 or 19) good catch !! Answer is C. Hi Bunuel, We (me and @sterling19) have come up with number substitution solution for this question but it took me 23 mins to write the test cases , is there a better way to attack this question apart from the magoosh official solution ?
_________________
Thanks, Lucky
_______________________________________________________ Kindly press the to appreciate my post !!



Manager
Joined: 09 Oct 2015
Posts: 95

The numbers D, N, and P are positive integers, such that D < N, and N [#permalink]
Show Tags
17 Nov 2015, 08:58
Bunuel Bunuel wrote: Bunuel wrote: The numbers D, N, and P are positive integers, such that D < N, and N is not a power of D. Is D a prime number?
(1) N has exactly four factors, and D is a factor of N (2) D = (3^P) + 2
Kudos for a correct solution. MAGOOSH OFFICIAL SOLUTION:This is a tricky one. Statement #1: two kinds of numbers have exactly four factors: (a) products of two distinct prime numbers, and (b) cubes of prime numbers. The product of two distinct prime numbers S and T would have factors {1, S, T, ST}. For example, the factors of 10 are (1, 2, 5, 10), and the factors of 21 are {1, 3, 7, 21}. The cube of a prime number S would have as factors 1, S, S squared, and S cubed. For example, 8 has factors {1, 2, 4, 8} and 27 has factors {1, 3, 9, 27}. We know N is not a power of D, so the second case is excluded. N must be the product of two distinct prime numbers. We know D < N, so of the four factors, D can’t be the product of the two prime numbers. D could be either of the prime number factors, or D could be 1, which is not a prime number. Because D could either be a prime number or 1, we cannot give a definitive answer to the question. This statement, alone and by itself, is not sufficient. Statement #2: this is tricky. The first few plugins seem to reveal a pattern. Even if you sense a pattern, it’s important to remember that plugging in numbers alone is never enough to establish that a DS statement is sufficient. Here, if we persevered to one more plugin, we would find the one that breaks the pattern. P = 5 > 3^5 + 2 = 243, which is not a prime. That gives another answer to the prompt, so we know this statement is not sufficient. To avoid a lot of plugging in, it’s also very good to know that in mathematics, prime numbers are notorious for not following any easy pattern. It is impossible to produce an algebraic formula that will always produce prime numbers. In fact, this is more than you need to know, but the hardest unsolved question in higher mathematics, the Riemann Hypothesis, concerns the pattern of prime numbers; mathematicians have been working on this since 1859, and no one has proven it yet. Suffice to say that no oneline algebraic formula is going to unlock the mystery of prime numbers! Combined statements: according to the information in statement #1, either D = 1 or D is a prime number. Well, statement #2 excludes the possibility that D = 1, because that number cannot be written as two more than a power of 3. Therefore, D must be a prime number. We have a definitive answer to the prompt question. Combined, the statements are sufficient. Answer = (C) if N is a cube of a prime, n = 27 in this case we can have d as either 1 or 9, right? why have we completely ignored the case of n being a cube of a prime no?



NonHuman User
Joined: 09 Sep 2013
Posts: 6643

Re: The numbers D, N, and P are positive integers, such that D < N, and N [#permalink]
Show Tags
12 Dec 2017, 05:39
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources




Re: The numbers D, N, and P are positive integers, such that D < N, and N
[#permalink]
12 Dec 2017, 05:39






